Why do many people call SSE the sum of squares of residuals? But the sum of squares of residuals in my econometrics book is RSS? What's the matter?

Why do many people call SSE the sum of squares of residuals? But the sum of squares of residuals in my econometrics book is RSS? What's the matter?

SSE (sum of squares for error) is the sum of squares of error terms, which reflects the error situation. RSS (residual sum of squares) also reflects the error situation. The formula is the same···

What does SSE mean in statistics and what is the calculation formula

SSE (sum & nbsp; of & nbsp; squares & nbsp; for & nbsp; error) is the sum of squares of residuals. It reflects the dispersion of each observation value of each sample, also known as the sum of squares within a group or the sum of squares of error terms

What is SSE! (Statistics)
R is equal to 0 only when the systematic error and random error of SSA are equal to 0 at the same time. However, if the random error is equal to 0, then SSE is equal to 0, and the square of denominator SST is equal to 0, then R is meaningless,
PS: it is clearly written in the book that SSA is a term about random error and systematic error, but if R is equal to 0, it is only true under the condition that "SSA is systematic error",

We should treat this problem dialectically
SSA is a term about random error and systematic error
So when the systematic error and random error of your SSA are equal to 0 at the same time, R is equal to 0
SSE will not be equal to 0

It is known that, as shown in the figure △ ABC, D, e and F are trilateral points respectively, and the area of △ DEF is 4cm & sup2;, so the area of △ ABC can be calculated?

The similarity ratio of △ def and △ ABC is 1 / 2
So: the area ratio is 1 / 4,
So: s △ ABC = 4S △ def = 4 * 4 = 16 (square centimeter)

The middle points of the three sides of △ ABC are D, e and f respectively. How many times of the area of △ DEF is the area of △ ABC?

First of all, I suggest you imagine (or draw a picture to see it clearly, so that you can understand it easily): suppose that △ ABC is an equilateral triangle, and the middle points of the three sides are D, e and f respectively, then the area of △ DEF is 1 / 4 of the area of △ ABC
The calculation method is as follows: [know Helen's area calculation formula: P = (a + B + C) / 2]
　S=√[p(p-a)(p-b)(p-c)]=（1/4）√[(a+b+c)(a+b-c)(a+c-b)(b+c-a)]　】
The lengths of the three sides are: A, B, C. the sides of △ def are: 1 / 2a, 1 / 2B, 1 / 2C
Using s = (1 / 4) √ [(a + B + C) (a + B-C) (a + C-B) (B + C-A), we can know that the area of △ DEF is 1 / 4 of the area of △ ABC
Hope to help you, if you have any questions can be added, happy to serve you!

If the area of △ ABC is 12cm ^ 2, and D, e and F are the midpoint of the three sides of △ ABC, then the area of △ def

3 square centimeters
The bottom and height of the triangle are half of the original
That's the reminder

In Δ ABC, points D, e and F are the midpoint of edges BC, AC and ab respectively. If s Δ ABC = 1.6, then the edge area of Δ DEF is
A.0.4 B.0.8 C.3.2 D.6.4

A: Let the length L of one side of triangle ABC and the corresponding height h, then its area is 1 / 2HL. According to ED parallel to AB, the area of triangle def = 1 / 2 * (1 / 2H) * (1 / 2L) = 1 / 4 (1 / 2HL) = 1 / 4 * 1.6 = 0.4

The three sides of △ ABC satisfy the equation 2A & sup2; + B & sup2; = 3AB + BC AC. let's ask the shape of △ ABC (2a-b + C ≠ 0)
I understand

2a²+b²-3ab-bc+ac=0
2a²+b²-2ab-bc+ac-ab=0
2a(a-b)+b(b-c)-a(b-c)=0
2a(a-b)+(b-c)(b-a)=0
(a-b)[2a-(b-c)]=0
A = B, or 2A + C = B (obviously impossible, because a + C > b)
So, a = B is isosceles

It is known that the three sides of △ ABC are a, B, C respectively, and | B + C-2A | + (B + C-5) 2 = 0. The value range of B is obtained

B + C-2A = 0, B + C-5 = 0, the solution is: B + C = 5, substituting B + C = 5 into B + C-2A = 0, the solution is: 5-2a = 0, the solution is: a = 2.5, then C = 5-b, according to the trilateral relationship of triangle: | 5-b-2.5 | B and B < 5-b + 2.5, that is, 2.5-b < B < 2.5 + 5-b, the solution is: 54 < B < 154

The three sides of the triangle are A.B.C, and the value of ABC satisfies the range of B by the equation B + C-2A = 0. B + C-5 = 0

Substituting B + C = 5 into B + C-2A = 0, a = 2.5,
If B-C = 2b-5 ＜ 2.5, B ＜ 3.75;
If C-B = 5-2B ＜ 2.5, B ＞ 1.25;
The range of B was 1.25 < B < 3.75